Problem 11 - Single Variable Calculus

Question

Determine whether the series is convergent, absolutely convergent, conditionally convergent, or divergent. \(\sum_{n=1}^{\infty} \frac{n !}{e^{n}}\)

Step-by-Step Solution

Answer

The series \(\sum_{n=1}^{\infty} \frac{n !}{e^{n}}\) is divergent, as determined by applying the ratio test and calculating the limit \(\lim_{n \to \infty} \frac{n+1}{e} = \infty\).

Step 1: Set up the ratio test

The ratio test requires that we find \(\lim_{n \to \infty} \frac{a_{n+1}}{a_n}\). So, we start by setting up this ratio: \(\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)!}{e^{n+1}}}{\frac{n!}{e^n}}\)

Step 2: Simplify the ratio

In this step, we'll simplify the expression by multiplying the numerator and the denominator by \(e^n * n!\): \(\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{e^{n+1}} * \frac{e^n * n!}{n! * e^n} = \frac{(n+1)!}{e^{n+1}} * \frac{n! * e^n}{n! * e^n}\) This simplifies to: \(\frac{a_{n+1}}{a_n} = \frac{(n+1)!e^n}{e^{n+1}n!}\)

Step 3: Further simplify the ratio

We can further simplify the expression by canceling out the common factors (\(e^n\) and \(n!\)): \(\frac{a_{n+1}}{a_n} = \frac{(n+1)!e^n}{e^{n+1}n!} = \frac{(n+1)n!}{e^{n+1}} * \frac{e^n}{n!} = \frac{(n+1)}{e}\)

Step 4: Calculate the limit

Now, we compute the limit \(\lim_{n \to \infty} \frac{(n+1)}{e}\): \(\lim_{n \to \infty} \frac{n + 1}{e} = \infty\) Since the limit is greater than 1, the ratio test indicates that the series diverges. Therefore, the given series \(\sum_{n=1}^{\infty} \frac{n !}{e^{n}}\) is divergent.

Key Concepts

Ratio Test

The ratio test is a commonly used method to determine the convergence or divergence of an infinite series. In the context of this problem, we use the ratio test to analyze the series \( \sum_{n=1}^{\infty} \frac{n!}{e^{n}} \). The test involves examining the limit \( \lim_{n \to \infty} \frac{a_{n+1}}{a_n} \), where \( a_n \) represents the general term of the series.

• If this limit is less than 1, the series is absolutely convergent.
• If it equals 1, the test is inconclusive.
• If it's greater than 1, the series is divergent.

In the solution steps, the ratio was set up and simplified, leading us to calculate the limit which turned out to be infinite. Since the limit is greater than 1, the series was determined to be divergent.

Infinite Series

An infinite series is the sum of the terms of an infinite sequence. It is expressed as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the terms in the sequence. The behavior of these series, either convergent or divergent, is crucial in calculus and mathematical analysis.
When an infinite series converges, its terms approach a particular value, contributing to a finite sum. Divergence, on the other hand, means that the sum of the series approaches infinity. The series \( \sum_{n=1}^{\infty} \frac{n!}{e^{n}} \) discussed here results in divergence due to the exponential growth of the factorial numerator compared to the exponential denominator.

Factorials

Factorials, denoted by \( n! \), represent the product of all positive integers up to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Used in the series \( \frac{n!}{e^n} \), factorials are significant in understanding why the series diverges.
As \( n \) increases, \( n! \) grows much faster than the exponential base \( e^n \). This rapid growth plays a key role in the eventual divergence, as the terms do not shrink fast enough to add to a finite sum.

Limits in Calculus

Calculus often uses limits to understand the behavior of functions and sequences as they approach certain points or infinity. In this problem, limits are crucial for applying the ratio test.
We calculate \( \lim_{n \to \infty} \frac{n+1}{e} \), which is fundamental in concluding that the series diverges. The limit describes how the terms \( \frac{n!}{e^n} \) behave as \( n \) becomes very large. If such a limit results in infinity or a value larger than 1, as seen here, it suggests the series does not converge to a finite sum.